# Law of cosines

## Branko Ćurgus

Theorem. If $ABC$ is a triangle with the angle $\gamma$ at the vertex $C$ and with the sides lengths $a = \overline{BC}, b= \overline{AC}$ and $c= \overline{AB}$ (see the figure below), then $c^2 = a^2 + b^2 - 2ab \cos \gamma.$
Proof. In the proof below we use the Pythagorean theorem twice and the definition of $\cos \gamma = x/a$;
 $\displaystyle c^2$ $\displaystyle = y^2 + h^2$ (by Pythagorean theorem $c^2 = y^2+h^2$) $\displaystyle = (b-x)^2 + h^2$ (by the identity $b = x + y$) $\displaystyle = b^2 -2bx + x^2 + h^2 \$ (algebra) $\displaystyle = a^2 + b^2 - 2bx$ (by Pythagorean theorem $a^2 = h^2 + x^2$) $\displaystyle = a^2 + b^2 - 2ab\dfrac{x}{a}$ (algebra) $\displaystyle = a^2 + b^2 - 2ab\cos \gamma$ (the definition of $\cos \gamma$)
This completes the proof.
Remark. Why is the law of cosines important in a linear algebra class? Here is why. Introduce two vectors in the above figure: $\vec{a} = \vec{CB}, \qquad \vec{b} = \vec{CA}.$ Then, $a^2 = \bigl\|\vec{a}\bigr\|^2$, $b^2 = \bigl\|\vec{b}\bigr\|^2$ and $\vec{b} - \vec{a} = \vec{BA}.$ The last equality implies \begin{align*} c^2 & = \| \vec{b} - \vec{a} \|^2 \\ & = \bigl(\vec{b} - \vec{a} \bigr) \cdot \bigl(\vec{b} - \vec{a} \bigr) \\ & = \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{b} + \vec{a} \vec{a} \\ & = \bigl\|\vec{b}\bigr\|^2 - 2 \vec{a} \cdot \vec{b} + \bigl\|\vec{a}\bigr\|^2 \\ & = a^2 + b^2 - 2 \vec{a} \cdot \vec{b}. \end{align*} Now compare the last equality and the law of cosines: \begin{align*} c^2 & = a^2 + b^2 - 2 \vec{a} \cdot \vec{b}, \\ c^2 &= a^2 + b^2 - 2ab \cos \gamma. \end{align*} We deduce that $\vec{a} \cdot \vec{b} = ab \cos \gamma = \bigl\|\vec{a}\bigr\| \, \bigl\|\vec{b}\bigr\| \cos \gamma.$ This gives us a geometric interpretation of the dot product.